Question

the sum of two numbers is 80 and the greater number exceeds twice the smaller number by 11 .find the number

Answer 1

x+y = 80.........equation first
x = 2y+ 11.......equation second
from second equation
x-2y = 11
x+y. = 80
subtract :
x -2y = 11
-x- y. = - 80
.......................
-3y = -69
y = 21



so ,
x+y = 80
put the value of y = 21
x + 21 = 80
x = 80 -21
x = 59



therefore, x = 59, y = 21

Answer 2

$\huge\underline\mathcal\orange{Answer}$

Let the number be x

Then greater number = 2x + 11

Sum of two numbers = 80

x + 2 + 11 = 80

3x = 80 - 11

3x = 69

x = 23

Hence, smaller number = 23

Greater number = 2 × 23 + 11 = 57